Question: A curve in the plane is defined parametrically by the equations $x=t^3-t$ and $y=\sqrt{3t+1}$. Find the value of $\dfrac{dy}{dx}$ at $t=1$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3}{4}$ (Choice B) B $\dfrac{3}{8}$ (Choice C) C $0$ (Choice D) D $\dfrac{1}{8}$
Explanation: In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=t^3-t$ and $y=\sqrt{3t+1}$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}(\sqrt{3t+1})}{\dfrac{d}{dt}(t^3-t)} \\\\ &=\dfrac{\left(\dfrac{3}{2\sqrt{3t+1}}\right)}{3t^2-1} \\\\ &=\dfrac{3}{2\sqrt{3t+1}(3t^2-1)} \gray{\text{Simplify}} \end{aligned}$ Now let's evaluate $\dfrac{dy}{dx}$ at $t= 1$ : $\begin{aligned} &\phantom{=}\dfrac{3}{2\sqrt{3( 1)+1}(3( 1)^2-1)} \\\\ &=\dfrac{3}{2\sqrt{4}(2)} \\\\ &=\dfrac{3}{8} \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $t=1$ is $\dfrac{3}{8}$.